The Stacks project

110.52 A torsor which is not an fppf torsor

In Groupoids, Remark 39.11.5 we raise the question whether any $G$-torsor is a $G$-torsor for the fppf topology. In this section we show that this is not always the case.

Let $k$ be a field. All schemes and stacks are over $k$ in what follows. Let $G \to \mathop{\mathrm{Spec}}(k)$ be the group scheme

where $\mu _{2, k}$ is the group scheme of second roots of unity over $\mathop{\mathrm{Spec}}(k)$, see Groupoids, Example 39.5.2. As an inverse limit of affine schemes we see that $G$ is an affine group scheme. In fact it is the spectrum of the ring $k[t_1, t_2, t_3, \ldots ]/(t_ i^2 - 1)$. The multiplication map $m : G \times _ k G \to G$ is on the algebra level given by $t_ i \mapsto t_ i \otimes t_ i$.

We claim that any $G$-torsor over $k$ is of the form

for certain $a_ i \in k^*$ and with $G$-action $G \times _ k P \to P$ given by $x_ i \to t_ i \otimes x_ i$ on the algebra level. We omit the proof. Actually for the example we only need that $P$ is a $G$-torsor which is clear since over $k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots )$ the scheme $P$ becomes isomorphic to $G$ in a $G$-equivariant manner. Note that $P$ is trivial if and only if $k' = k$ since if $P$ has a $k$-rational point then all of the $a_ i$ are squares.

We claim that $P$ is an fppf torsor if and only if the field extension $k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots )/k$ is finite. If $k'$ is finite over $k$, then $\{ \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)\}$ is an fppf covering which trivializes $P$ and we see that $P$ is indeed an fppf torsor. Conversely, suppose that $P$ is an fppf $G$-torsor. This means that there exists an fppf covering $\{ S_ i \to \mathop{\mathrm{Spec}}(k)\}$ such that each $P_{S_ i}$ is trivial. Pick an $i$ such that $S_ i$ is not empty. Let $s \in S_ i$ be a closed point. By Varieties, Lemma 33.14.1 the field extension $\kappa (s)/k$ is finite, and by construction $P_{\kappa (s)}$ has a $\kappa (s)$-rational point. Thus we see that $k \subset k' \subset \kappa (s)$ and $k'$ is finite over $k$.

To get an explicit example take $k = \mathbf{Q}$ and $a_ i = i$ for example (or $a_ i$ is the $i$th prime if you like).